Question

The solution of the inequality 2a < a – 4 ≤ 3a + 8 is

Answer 1

$\large\underline{\sf{Solution-}}$

Given linear inequality is

$\rm :\longmapsto\:2a < a - 4 \leqslant 3a + 8$

Consider,

$\red{\rm :\longmapsto\:2a < a - 4}$

On Subtracting 'a' from both sides, we get

$\red{\rm :\longmapsto\:2a - a < a - a - 4}$

$\red{\rm :\longmapsto\:a < - 4 - - - - (1)}$

Now, Consider,

$\purple{\rm :\longmapsto\:a - 4 \leqslant 3a + 8}$

On Subtracting 3a from both sides, we get

$\purple{\rm :\longmapsto\:a - 4 - 3a \leqslant 3a + 8 - 3a}$

$\purple{\rm :\longmapsto\: - 4 - 2a \leqslant 8}$

On Adding 4 both sides, we get

$\purple{\rm :\longmapsto\: - 4 - 2a + 4 \leqslant 8 + 4}$

$\purple{\rm :\longmapsto\: - 2a \leqslant 12}$

$\purple{\rm :\longmapsto\: a \geqslant - 6 - - - - (2)}$

From equation (1) and (2), we concluded that

$\bf\implies \: - 6 \leqslant a < - 4$

$\bf\implies \:a \: \in \: [ \: - \: 6, \: - \: 4)$

Additional Information :-

$\red{ \boxed{ \sf{ \:x > - y \: \: \implies \: - x < y}}}$

$\blue{ \boxed{ \sf{ \:x < - y \: \: \implies \: - x > y}}}$

$\green{ \boxed{ \sf{ \:x \leqslant - y \: \: \implies \: - x \geqslant y}}}$

$\pink{ \boxed{ \sf{ \:x \geqslant - y \: \: \implies \: - x \leqslant y}}}$

$\red{ \boxed{ \sf{ \: |x| < y \: \: \implies \: - y < x < y}}}$

$\blue{ \boxed{ \sf{ \: |x| \leqslant y \: \: \implies \: - y \leqslant x \leqslant y}}}$

$\green{ \boxed{ \sf{ \: |x| \geqslant y \: \: \implies \: - y \geqslant x \geqslant y}}}$